PAT 1023. Have Fun with Numbers (20) 模拟大数乘法

#include<cmath>#include<algorithm>#include<cstring>#include<iostream>#include<stack>#include<vector>#include<queue>#include<string>#include<map>using namespace std;#define M 30int inum[M];int ans[M];//90-48=42min//耗时于各种错误，变量搞错，参数搞混，以及num>=10写成num>10//题意：一个数字，*2之后，是否是原数字的排列组合//我的做法：字符串存入数组，模拟乘2，存入ans数组，并对输入和ans作排序//排序之后就可以一一比较，全部一致则为Yes//注意，因为ans已经被排序了，所以要弄一个副本ansputint main(){int add,i,j;string str;    cin>>str;int n = str.size();add=0;i=0;int ansput[M];for(int ii=n-1;ii>=0;ii--){int num=str[ii]-'0';inum[i]=str[ii]-'0';num *= 2;num += add;if(num >= 10){add = 1;num -= 10;}else add = 0;ansput[i] = num;ans[i++] = num;}if(add==1){ans[i++]=1;printf("No\n");for(j=n;j>=0;j--){cout<<ans[j];}cout<<endl;return 0;}sort(inum,inum+n);sort(ans,ans+n);for(i=0;i<n;i++){if(inum[i]!=ans[i]){printf("No\n");for(j=n-1;j>=0;j--){cout<<ansput[j];}cout<<endl;return 0;}}printf("Yes\n");for(j=n-1;j>=0;j--){cout<<ansput[j];}cout<<endl;    return 0;}